# Global maximum and minimum

• Dec 7th 2011, 08:34 PM
Happytime
Global maximum and minimum
I am having trouble with finding the global maximum and minimum values of this function:

$\displaystyle \frac {1} {(x-8)^2+6}$

Am I supposed to be finding the derivative of it and then setting it to zero or I'm I thinking the completely wrong thing here?
• Dec 7th 2011, 08:37 PM
pickslides
Re: Global maximum and minimum
You are on the right track, I'd sketch the function first, maybe that will give you some hints.
• Dec 8th 2011, 12:35 AM
sbhatnagar
Re: Global maximum and minimum
Quote:

Find global minimum and maximum of $\displaystyle \frac{1}{(x-8)^2+6}$
Let $\displaystyle f(x)=(x-8)^2+6$ and $\displaystyle g(x)=\frac{1}{(x-8)^2+6}$

You must understand that:

$\displaystyle \max{\left\{\ g(x)\right\}}=\frac{1}{\min{\left\{ f(x)\right \}}}$ and $\displaystyle \min{\left\{\ g(x)\right\}}=\frac{1}{\max{\left\{ f(x)\right \}}}$

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$\displaystyle \min{\left\{ (x-8)^2+6\right\}}=6 \ \text{{\color{red}at}} \ x=8$...<------Why? Find out by using your knowledge of maxima and minima.

$\displaystyle \max{\left\{ \frac{1}{(x-8)^2+6}\right\}}=\frac{1}{6} \ \text{{\color{red}at}} \ x=8$